Answer:74.75 kJ/kg
Explanation:
Given
mass of hot fluid[tex](m_h)[/tex]=5 kg/s
Energy associated with it=150 kJ/kg
mass of cold fluid[tex](m_l)[/tex]=15 kg/s
Energy associated with it=50 kJ/kg
Energy lost=5.5 kW
Let the Enthalpy of outlet fluid be h
mass at outlet
[tex]m=m_l+m_h=20 kg/s[/tex]
Using Energy Conservation
[tex]m_h\times E_h+m_l\times E_l+Q=m\times h[/tex]
[tex]5\times 150+15\times 50-5.5=20\times h[/tex]
h=74.75 kJ/kg